Posted by **anisah** on Sunday, April 22, 2012 at 3:44am.

a steel ball weighing 128 lb is suspended from a spring, whereupon the spring is stretched 2 ft from its natural length. the ball is started in motion with no initial velocity by displacing it 6 in above the equilibrium position. assuming no damping force, find an expression for

a)the position of the ball at any time.

b)the position of the bal at t=pai/12 sec

c)the circular frequency, natural frequency and period for this system

- differential equations -
**MathMate**, Sunday, April 22, 2012 at 9:41am
This is a "spring-mass" mass problem.

m=128 lb

L=2 ft

g=32.2 ft/sec² (accel. due to gravity)

k=mg/L=128*32.2/2=2060.8

Let y=displacement (downwards) in the same direction as g, then the differential equation that governs the motion is:

acceleration = -ky, or

y"+ky=0 where y"=d²y/dt²

The solution of the auxiliary equation is

(m^2+k)=0, or m=±i√k

and the solution to the equation is

y=C1 cos(√k t) + C2 sin(√k t)

The initial conditions are:

y(0)=-0.5, y'(0)=0.

Since

y(0)=-0.5=C1 cos(0) + C2 sin(0)

C2=0, and C1=-0.5

y'(t)=-C2√k sin(√k t) + C2 √k cos(√k t)

From

y'(0)=0=-C2√k sin(0) + C2 √k cos(0)

we conclude

C2=0 since sin(0)=0, √k≠0 and cos(0)≠0.

So the equation of motion is:

y=-(1/2)cos(√ t)

Answers to (b) and (c) can be deduced from the equation of motion.

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